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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Why my code did not work?

I have tried this code on an online compiler and my code worked. However when I use it for the exercise, it does not work.

wordcount.py
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.
def word_count(st):
    dic = {}
    copy = {}
    arr = st.split()
    for word in arr:
        if len(dic) == 0:
            dic[word.lower()] = 1
            continue
        for new in dic:
            copy[new] = dic[new]
        for key in dic:
            if key == word.lower():
                copy[key] += 1
                break
            else:
                copy[word.lower()] = 1
        for new_1 in copy:
            dic[new_1] = copy[new_1]
    return dic
Istvan Nonn
Istvan Nonn
2,092 Points

For sure your code gives back the expected result, however I will highly suggest to rewrite your code. It is good as a first approach, but than just try to clean it up and think if you can make it shorter. There are way too many for loops and if statements.

1 Answer

Istvan Nonn
Istvan Nonn
2,092 Points

I am not sure (just yet) how the code checker works, but most like there is a pattern or a strict way / order the code needs to be written. Anyway, I'll keep looking around for other solutions, meanwhile if anyone can help to solve the mystery will be much appreciated:

def word_count(sentence):
    sentence = sentence.lower().split(" ")
    words = {}
    for word in sentence:
        number_of_times = sentence.count(word)
        words[word] = number_of_times
    return words